CHAPTER 12 SOLUTIONS TO EXERCISES ON COMPLEX NUMBERS
Exercise on 12.1
Solve the following equations in a + jb form:-
i)x2 + 4 = 0ii)x2 + x + 1 = 0iii)x2 + 6x + 11 = 0
iv)x3 – 1 = 0
Solutions
i)x2 + 4 = 0
We have
x2 = – 4 or x = = = 2j
ii)x2 + x + 1 = 0
Using the formula for the solution of a quadratic we have
x = = = – ± j
iii) x2 + 6x + 11 = 0
x = = = – 3 ± j
iv)x3 – 1 = 0
Here we first factorize the cubic to give
x3 – 1 = (x – 1)(x2 + x + 1) = 0
This gives solutions, from ii)
x = 1, – ± j
Exercise on 12.2
For the complex numbers z = 3 – j and w = 1 + 2j evaluate
i) 2z – 3wii) zwiii)z2w 2iv)
Solutions
i) 2z – 3w = 2(3 – j) – 3(1 + 2j) = 6 – 2j – 3 – 6j = 3 – 8j
ii) zw = (3 – j)(1 + 2j)= 3 – j + 6j – 2 j2 = 3 + 5j – 2(– 1) = 5 + 5j
on remembering that j2 = – 1
iii)Don't be so hasty with z2w 2 – look before you leap and notice that
z2w 2 = w(z )2 = (1 + 2j)= 10(1 + 2j) = 10 + 20j
iv)= = = =
on multiplying top and bottom by the conjugate of 1 + 2j
Exercises on 12.3
1.Plot the following complex numbers on the Argand plane and put them into polar form.
i) 1 ii) j iii) – 3j iv) 1 – j v) 2 + j vi) – 3 – 2j viii) – 3 + 2j
Solution
i) 1 has modulus r = = 1 and as it lies on the positive real axis the principal value of its argument is 0 radians, and so
1 = 1(0)
ii) |j| = 1 and j lies on the imaginary axis so its argument is = . So
j = 1
iii) |– 3j| = 3 and the PV of the argument is = – , so
–3 j = 3
iv) |1 – j| = so
1 – j =
=
=
=
So
1 – j =
v) 2 + j = = () where tan =
vi) – 3 – 2j = = () where tan = + tan–1
Here the complex number is in the third quadrant and we have chosen the angle in the range 0, 2.
viii) – 3 + 2j = = () where tan = – tan–1
2.Convert into Cartesian form
i) 2 (0)ii) 3()iii) (/2)
iv) 3v) 1vi) 2(/2)
Solution
i) 2 (0) = 2(cos(0) + j sin (0)) = 2
ii) 3() = 2(cos() + j sin ()) = 2 (– 1 + j 0) = – 2
iii) (/2) =cos(/2) + j sin (/2) = 0 + j 1 = j
iv) 3= 3= 3= – + j
v) 1= 1= cos – j sin
=– j
vi) 2(/2) = –2j (here it is as quick to think about the Argand diagram).
Exercises on 12.4
1. Evaluate all possible products, excluding powers, of the three complex numbers below and compare with the results obtained in a + jb form (See Q2 Exercises on 12.3).
z1 = (/2) z2 = 3 z3 = 1
Solution
z1 z2 = (/2)3= 3= 3
= 3=
Compare this with
(/2)3= j =
z1 z3 = (/2) 1= = =+ j
Compare this with
(/2) 1= j= + j
z2 z3 = 31= 3 =
= 3 cos + 3 sinj
Compare this with
31=
= + j
which shows that
cos = and sin =
z1 z2 z3 = (/2)31= 3= 3
= 3 cos + 3 sin j
Compare this with
z1 z2 z3 = j= j
= – + j
which shows that
cos = – and sin =
2.Show that if z = r () , then
z2 = r2(2) , z3 = r3(3)
What do you think is the result for zn where n is a positive integer? You will see more of this in Section 12.6
Solution
Using r1(1) r2(2) = r1 r2(1 + 2) with 1 = 2 = we get
z2 = r2() () = r2() = r2(2)
Then
z3 = z2 z = r3(2) () = r3(2 + ) = r3(3)
It does not take a lot of imagination to extrapolate this to
zn = rn(n)
Exercise on 12.5
If z1 = and z2 = evaluate in polar and Cartesian form
Solution
In polar form:
= = =
= cos + sin j
In Cartesian form:
= =
= – + j = + j
from which
cos = and sin =
Exercises on 12.6
1.Find the modulus of ej
Solution
|ej| |cos + j sin | =
2.Show that (ej)–1 = e–j
Solution
(ej)–1 = (cos + j sin )–1 =
= = cos – j sin = cos (– + j sin (– ) = e–j
3.Show that cos = sin=
Solution
We have
ej = cos + j sin
e–j = cos – j sin
Adding these identities gives
2 cos = ej + e–j so cos =
Subtracting the results gives
2 j sin = ej – e–j so sin =
Exercise on 12.7
By conversion to polar form and use of De Moivres' theorem evaluate
i)j7ii)(1 + j)5iii)(– j)–4
Solution
i) j7 = 7 = = = – j
ii) (1 + j)5 = 5 = 55 = 4
= – 4= – 4(1 + j)
iii) First, we have – j = 2= 2 = 2
So
(– j)–4 = –4 = 2 –4= 2– 4
Exercises on 12.8
1.Find in a + jb form and plot on Argand diagram:-
i)The three values of j1/3
ii)The four values of (1 + j)1/4
Solution
i) j1/3 = =
= for k = 0, 1, 2
k = 0 gives
= = cos + j sin =+ j
k = 1 gives
= = cos + j sin = – + j
k = 2 gives
= = – j
ii) (1 + j)1/4 =
= 2= 2
= 2for k = 0, 1, 2, 3
k = 0 gives
= 2= 21/8
k = 1 gives
= 2= 2
= 21/8
k = 2 gives
2= 2
= 21/8
k = 3 gives
2= 2
= 21/8
2.Solve the following equations in a + jb form and plot the roots in the Argand plane.
i)x6 – 1 = 0ii) x3 + 8 = 0
Solution
i)x6 – 1 = 0 or x6 = 1 so
x = ((0))= k = 0, 1, ... , 5
=
= k = 0, 1, ... , 5
k = 0 gives
= (0) = 1
k = 1 gives
= cos + j sin =+ j
k = 2 gives
= cos + j sin = – + j
k = 3 gives
() = –1
k = 4 gives
= cos + j sin = –– j
k = 5 gives
= cos + j sin = – j
ii) x3 + 8 = 0 is equivalent to
x3 = – 8 = 8 () = 8 ( + 2k)
So
x = 2
= 2for k = 0, 1, 2
k = 0 gives
= 2 = 2 = 1 + j
k = 1 gives
= 2 () = – 2
k = 2 gives
= 2 = 2 = 1 – j
REINFORCEMENT EXERCISES IN COMPLEX NUMBERS
1.Write the following in simplest form in terms of real numbers and j.
i)ii)iii)
iv)j2v)– j2vi)
viii)(– j)2
Solution
i)= j of course
ii)= 3 (by convention gives the positive root)
iii)= 3= 3j
iv)j2 = –1
v)– j2 = – (– 1) = 1
vi) The easiest way to get at is to divide j2 = –1 by j to get:
= – j
viii) (– j)2 = j2 = –1
2.Solve the following equations, writing the answer in z = a + jb form
i)z2 + 25 = 0ii)z2 + 4z + 5 = 0
iii)z4 – 3z2 – 4 = 0iv)z3 + z – 2 = 0
v)z3 + 1 = 0vi)z2 + 2jz + 1 = 0
Using egs (i) – (v) verify the result that the equations with real coefficients have real roots and/or complex roots occurring in conjugate pairs.
Solution
i)z2 + 25 = 0gives z2 = – 25 so
z = = = 5= 5j
ii)z2 + 4z + 5 = 0 gives, using the quadratic formula,
z = =
= = – 2 j
iii) z4 – 3z2 – 4 = 0 is really a quadratic equation in disguise, because if we put u = z2 then we get
z4 – 3z2 – 4 = u2 – 3u – 4 = (u + 1)(u – 4) = 0
So u = –1 or z2 = – 1 and hence z = j or u = 4 = z2 so z = 2
iv) To solve z3 + z – 2 = 0 we first factorize it. By the factor theorem, the fact that z = 1 is (by trial) a root tells us that z – 1 is a factor, so we can write
z3 + z – 2 (z – 1)(z2 + az + b)
Equating coefficients gives b = 2 and – a + b = 1, so a = 1 and hence
z3 + z – 2 (z – 1)(z2 + z + 2) = 0
Thus, z = 1 or z2 + z + 2 = 0 ie
z = =
v) Using the factorization of a sum of cubes we have
z3 + 1 (z + 1)(z2 – z + 1) = 0
and so
z = – 1 or z = =
vi) Even though the equation z2 + 2jz + 1 = 0 has complex coefficients we can still use the usual quadratic formula to obtain the solution:
z = =
= (–1 )j
The roots of all equations i) to v) are either real or occur in conjugate pairs – for example – 2 + j and –2 – j. The roots of vi) are not conjugate pairs because the coefficients of the equation are not all real.
3.Express in the form a + jb
i)j3ii)j27iii)3(1 + j) – 2(1 – j)
iv)(– 2j)6v)j(j + 2)vi)j/j
vii)2j(j – 1) + j3(2 + j)
Solution
i)j3 = j2 j = – 1 j = – j
ii)j27 = j26 j = 13 j = 13 j = – j
iii)3(1 + j) – 2(1 – j) = 3 + 3j – 2 + 2j = 1 + 5j
iv) (– 2j)6 = (– 2)6 j6 = 26 (j2)3 = 64 (– 1)3 = – 64
v)j(j + 2) = j2 + 2j = – 1 + 2j
vi) j/j = j2 = – 1
vii) 2j(j – 1) + j3(2 + j) = 2j2 – 2j – j(2 + j) = – 2 – 2j – 2j – j2
= – 2 – 4j + 1 = – 1 – 4j
4.Find the real and imaginary parts of :
i)(1 – j) (1 + j)ii)(3 – 4j) (1 + j) iii) (4 + 3j)2 iv) (4 + 3j)3
Solution
i)(1 – j) (1 + j) = 1 – j2 = 1 – (– 1) = 2
So real part = 2, imaginary part = 0
ii) (3 – 4j) (1 + j) = 3 + 3j – 4j – 4j2 = 3 – j + 4 = 7 – j
So real part = 7, imaginary part = –1
iii) (4 + 3j)2 = 42 + 24j +( 3j)2 = 16 + 24j – 9 = 7 + 24j
So real part = 7, imaginary part = 24
iv) By the binomial theorem
(4 + 3j)3 = 43 + 3. 42. 3j + 3.4. (3j)2 + (3j)3
= 64 + 6 . 16j – 4. 27 – 27j = –44 + 69j
So real part = – 44, imaginary part = 69
5.Write down the complex conjugates of:
i)5 + 3jii) iii)iv)
v)4vi)
Where appropriate put both the original complex number and its conjugate into a + jb form and check your results.
Note that you are not asked to write the complex conjugates in the form a + jb, but simply to state the complex conjugate of each given number. For this you only have to change the sign of j wherever it occurs. Then put the results in a + jb form and check against the original number in a + jb form.
Solution
i)If z = 5 + 3j then we get immediately = 5 + 3 (– j) = 5 – 3j
ii) If z = then = =
Now
= = + j
while
= = – j
which is just the complex conjugate of + j as required.
iii) If z = then = =
Now
= = + j
while
= = – j the complex conjugate of + j as required.
iv) If z = then = = =
= – – j
You can confirm that z = = – + j and so obtain = – – j directly as a check.
v)With z = 4 we have = 4 = 4
We know from iv) that = – + j and so
4 = 4
and so it is obvious directly that
= 4 = 4
without having to expand to check.
vi) With z = we have =
Now
= = + j
While
= = – j
as required. Note that in fact we could have worked entirely with the reciprocals in this case, without explicitly converting to a + jb form.
6.Put into a + jb form
i)(2 + 5j)(4 – 3j)ii)(4 – j)(1 + j)(3 + 4j)iii) iv) iv)
Solution
i)(2 + 5j)(4 – 3j) = 8 – 6j + 2j – 15j2 = 8 + 14j + 15 = 23 + 14j
ii)(4 – j)(1 + j)(3 + 4j) = (4 – j)(3 + 7j – 4) = (4 – j)(– 1 + 7j)
= – 4 +28j + j + 7 = 3 + 29j
Note how higher degree products can always be reduced to quadratic multiplication in complex numbers.
iii) = = =+ j
iv) = = = = – – j
iv) In this case it helps to work out the denominator first
= =
= = = – + j
7.Evaluate
i)+ ii)– iii)
and explain why each is either purely real or purely imaginary.
Solution
i) z = + = on taking the common denominator
= =
This is purely real, as we would expect from the original form which is clearly its own complex conjugate, z = . In fact z is of the form z = w + = 2 Re (w).
ii) z = – =
= = =j
From the original form we see that z = – and so – is purely imaginary, as we have found from the calculation. In fact z is of the form z = w – = 2 Im (w) j.
iii) z = = =which confirms that z is real as is clear from the original form. In fact z is of the form = and this is real and positive.
8.Simplify the complex number + . Find the modulus and argument of the result.
Solution
z = + =
= = = =
= =+ j
For the modulus we have
|z| = = =
The argument is seen to be from the figure.
9.State by inspection only (no arithmetic is necessary) whether each of the following numbers is purely real, purely imaginary or complex.
i)– ii)– iii)34
Solution
i)If z = – then we have
= – = – = – z
So z is pure imaginary.
ii) If z = – then
= – = – = z
So z is real.
iii) If z = 34 then
= 34
This has no simple relation to z and is certainly not equal to z, so in this case z is neither real nor pure imaginary.
10.Mark each of the following numbers on an Argand diagram and find the modulus and the principal value of the argument of each:
i)2ii)– 1iii)3j
iv)– jv)1 + jvi)– – j
vii)– 2 + 2jviii)– 3 – 3j
Write down the numbers in polar form.
Solution
i)For z = 2, |z| = 2 and Arg z = 0, so 2 = 2 (0)
ii) For z = – 1, |z| = 1 and Arg z = , so – 1 = 1 ()
iii)For z = 3j, |z| = 3 and Arg z = so 3j = 3
iv)For z = – j, |z| = 1 and Arg z = – so – j = 1
v)For z = 1+ j
|z| = = = 2
and
Arg z = Arg (1 +j) = tan– 1in the first quadrant.
Using the 60-30 triangle tells us that
Arg (1 + j) =
and so
1 + j = 2
vi)For z = – – j, |z| = 2 and Arg z = – from the figure
so
– – j = 2
vii)For z = – 2 + 2j
|z| = = = 2
and
Arg z = Arg(– 2 + 2j) =
so
– 2 + 2j = 2
viii) For z = – 3 – 3j,
|z| = = 3
and
Arg z = Arg (– 3 – 3j) = –
so
– 3 – 3j = 3
11.Convert to Cartesian form
i)4(0)ii)3iii)2 ()
iv)10()v)10vi)2
vii)3viii)2ix)
x)3xi)xii)3
Solution
i)4(0)4(cos 0 + j sin 0) = 4
ii) 3= 3= – 3 j sin= –3j
iii)2() = 2(cos + j sin ) = 2(– 1) = – 2
iv)10() = 10(cos + j sin ) = 10(– 1) = – 10
v) 10= 10= 10 j sin= 10j
vi)2= 2= 2= + j
vii) 3= 3= 3=– j
viii) 2= 2= 2
= 2= –– j
ix)= = =+ j
x)3= 3= 3
= 3= – – j
xi)= cos+ j sin = + j
xii)3= 3= 3
= 3= – – j
Notice how much easier it is to use surds (UEM 20) rather than their decimal approximations.
12.If = 5, Arg z1 = /3, = 3, Arg z2 = /4, find the Cartesian forms of z1 and z2 and the values of:
i)ii)iii)
iv)Arg (z1z2)v)Arg vi)Arg 1
Solution
Remember, for multiplication in polar form:
Multiply moduli
Add arguments
For division in polar form:
Divide moduli
Subtract arguments
Given and Arg z1 we can write z1as z1= ( Arg z1).
So
z1= 5 = 5= 5
Similarly
z2= 3 = 3= 3
i) = = 5 3 = 15
ii) = =
iii)= 2 = 52 = 25
iv)Arg (z1z2) = Arg (z1) + Arg (z2) =+ =
v) Arg = Arg (z1) – Arg (z2) =– =
vi)Arg 1 = – Arg z1 = –
13.Show that multiplication by j rotates a complex number through in the anticlockwise direction and division by j rotates it in the clockwise direction.
Solution
Let z = r()be any complex number, with modulus r and argument (not necessarily the principal value). Since j = 1 we have
jz = 1r()r
which shows that the argument of jz is + - ie rotated anti-clockwise through .
On the other hand
= = r
So the argument of is – , ie rotated clockwise through .
14.If z1 = 3,z2 = 2,z3 =
Find the polar form of
i)z1 z2ii)iii)z1 z2 z3
iv)v)vi)
Solution
We have to take all the angles in consistent form, and perhaps degrees are most convenient, as we don't have to deal with fractions then. So write: z1 = 3(30)z2 = 2(10)z3 = (60). Then
i) z1 z2 = 3(30) 2(10) 6(3010)6(40)
ii) = = (30– 10) = (20)
iii) z1 z2 z3 = 3(30) 2(10)(60) = 6(301060)6(100)
iv)We have z= 2 = (120) and so
= = 2(10 – 120)2( – 110)
v) = = (1060– 30)
= (40)
vi) z= 2 = 9 (60) so
=
= (6060–10) = (110)
15.Evaluate the powers indicated by use of the polar form.
i)(1 – j)8ii)(+ j)6iii)(2 + 2j)4
Solution
i)1 – j = =
So
(1 – j)8 = 8 = 24= 16 (– 2) = 16
ii) + j = 2= 2
So
(+ j)6 = 6 = 26() = – 64
iii)(2 + 2j)4 = 24(1 + j)4 and
1 + j = =
So
(2 + 2j)4 = 244 = 26 () = – 64
16.Plot the complex numbers
i)1 – jii)2jiii)+ j
iv)– 3j
on the Argand diagram and put them in the form ej
Solution
i)= and Arg (1 – j) = – , so
1 – j = e– j/4
ii) = 2 and Arg (2j) = , so
2j = 2ej/2
iii) = 2 and Arg = , so
+ j = 2ej/6
iv) = 3 and Arg (– 3j) = – , so
– 3j = 3e– j/2
Express the following numbers in the form a + jb:
i)ej/3ii)e–j/6iii)e–(1+j)/3iv)
v)vi)
Solution
i)ej/3 = cos + j sin = + j
ii) e–j/6 = cos + j sin = cos – j sin
= – j
iii) e–(1+j)/3 = e–1/3 e– j/3 = e–1/3
= e–1/3
= e–1/3= – j
iv) = = – j = – j – j2 =– j
v) = = =
= = =
=+ j
vi) = = =
=– j
18.Use the power series for ex with x = jto find ej in the form A + jB where A and B are real power series in . Hence show that:-
ej = cos+ jsin
Use this result to prove that
(cos+ jsin)n = cosn+ jsinn
Hence evaluate 43
Solution
The power series for exis
ex = 1 + x + + + … + + …
If we substitute x = j into this we get
ej = 1 + j + + + … + + …
= 1 + j – – j+ + …
on using j2 = – 1 repeatedly. Gathering real and imaginary terms together then gives
ej = 1 – + + … + j
Now we use the series for cosine and sine (remember is in radians)
cos = 1 – + + … and sin = – + …
to obtain the required relation:
ej = cos + j sin
Using this we have
as required.
Now
+ j = = ej/3
so, noting that 43 = 42 + 1 = 6 × 7 + 1
43 = 43 = 42
= 7 = ()7
=+ j
19.Simplify i) (1 + j )6 + (1 – j )6 and ii) (– j)15 by using De Moivre's theorem.
Solution
i) First put 1 + jand 1 – jinto polar form:
1 + j= 2= 2
and similarly:
1 – j= 2= 2
So
6 + 6 = 6 + 6
= 26(2) + 26(– 2) = 26 + 2627
ii) First put – j into polar form:
– j = 2= 2
= 2= 2
So
15 = 15 = 215
= 215= 215
= 215= – 215 j
20.Simplify
Solution
=
= = cos (– 38) + jsin (– 38)
= cos (38) – jsin (38)
21.Determine all the roots in each case
i) Square roots of 1ii)Square roots of – 1 iii) Square roots of j
iv) Cube roots of – j v) Square roots of (1 + j) vi) Square roots of 1 + j vii) Fourth roots of 1 – j viii) Fourth roots of + j
Solution
i) 1 = (0) = (0 + 2k) = (2k)
So
11/2 = ((2k))1/2 = (k) with k = 0, 1
= (0) or ()
= 1 or – 1
as we expect.
ii) (– 1)1/2 = (())1/2 = (( + 2k))1/2
= with k = 0, 1
= or
= j or – j
again as we expect.
iii) j1/2 = 1/2 = 1/2
= for k = 0, 1
= or
= cos + j sin =+ j
or
= cos + j sin = – – j
iv) ( – j)1/3 = 1/3 = 1/3
= for k = 0, 1, 2
For k = 0 we get
= = cos + j sin =– j
For k = 1 we get
= = = j (Check: j3 = j2 j = – j)
For k = 2 we get
= =
= cos + j sin = – – j
v) (1 + j)1/2 = 1/2 = 21/41/2
= 21/4for k = 0, 1
For k = 0 we get
= 21/4= 21/4
For k = 1 we get
= 21/4= 21/4
= 21/4
vi) (1 + j)1/2 = 1/2 = 21/21/2
= for k = 0, 1
For k = 0 we get
= =
=
For k = 1 we get
= =
=
We can summarise the two answers as:
vii) (1 – j)1/4 = 1/4 = 21/81/4
= 21/8for k = 0, 1, 2, 3
For k = 0 we get
= 21/8= 21/8
For k = 1 we get
= 21/8= 21/8
For k = 2 we get
= 21/8= 21/8
For k = 3 we get
= 21/8= 21/8
= 21/8
= 21/8
viii) (+ j)1/4 = 1/4 = 21/41/4
= 21/4for k = 0, 1, 2, 3
For k = 0 we get
= 21/4= 21/4
For k = 1 we get
= 21/4= 21/4
For k = 2 we get
= 21/4= 21/4
= 21/4
= 21/4
For k = 3 we get
= 21/4= 21/4
= 21/4
= 21/4
22.Find the values of z satisfying
z4 + 1 = 0.
Solution
We have
z4 = – 1 = ()
so
z = 1/4 = 1/4
= for k = 0, 1, 2, 3
k = 0 gives
=+ j
k = 1 gives
= –+ j
k = 2 gives
= –– j
k = 3 gives
=– j
23.Simplify each of the following numbers to the a + jb form:-
i)ii)iii)
Solution
i) =
=
= = cos7 + jsin7
ii)= = sin2 – jcos2
iii) =
= = sin – j cos
24.Find the polar forms of the fourth roots of – 16 and indicate the results on the Argand diagram.
Solution
(Compare with the solution to Exercise 22)
– 16 = 16(– 1) = 16 ()
so
(– 16)1/4 = 1/4 = 161/41/4
= 2 for k = 0, 1, 2, 3
k = 0 gives
2= 2=+ j
k = 1 gives
2=2= – + j
k = 2 gives
2= 2= – – j
k = 3 gives
2= 2= – j
– 1 –